What is the distribution of maximum of a pair of iid draws, where the minimum is an order statistic of other minima?

Question

Consider $n\cdot m$ independent draws from cdf $F(x)$, which is defined over 0-1, where $n$ and $m$ are integers. Arbitrarily group the draws into $n$ groups with m values in each group. Look at the minimum value in each group. Take the group that has the greatest of these minima. Now, what is the distribution that defines the maximum value in that group? More generally, what is the distribution for the $j$-th order statistic of $m$ draws of $F(x)$, where the kth order of those m draws is also the pth order of the n draws of that kth order statistic?

All of that is at the most abstract, so here is a more concrete example. Consider 8 draws of $F(x)$. Group them into 4 pairs of 2. Compare the minimum value in each pair. Select the pair with the highest of these 4 minima. Label that draw "a". Label the other value in that same pair as "b". What is the distribution $F_b(b)$? We know $b>a$. We know a is the maximum of 4 minimums of $F(x)$, of $F_a(a) = (1-(1-F(x))^2)^4$. What is $F_b(b)$?

Answer 1

I answer this: "Arbitrarily group the draws into n groups with m values in each group. Look at the minimum value in each group. Take the group that has the greatest of these minima. Now, what is the distribution that defines the maximum value in that group?"
Let $X_{i,j}$ the i-th random variable in group j and $f(x_{i,j})$ ($F(x_{i,j})$) its density (cdf) function.
Let $X_{\max,j}, X_{\min,j}$ the maximum and minimum in group $j$. Let $X_{final}$ the variable that results at the end of all process. We want to calculate $P(X_{final}<x)$ which is
$$P(X_{\max,j_0}<x \hbox{ and } X_{\min,j_0}=\max_j{X_{\min,j}} \hbox { and } 1\leq j_0\leq n)$$ $$=nP(X_{max,1}<x \hbox{ and } X_{\min,1}=\max_j{X_{\min,j}})$$ $$=nmP(X_{1,1}<x\hbox{ and } X_{1,1}=\max_i(X_{i,1})\hbox{ and } X_{\min,1}=\max_j{X_{\min,j}})$$ $$=nmP(X_{1,1}<x, X_{1,1}>X_{2,1}>\max_{j=2\ldots n} X_{min,j},\ldots,X_{1,1}>X_{m,1}>\max_{j=2\ldots n} X_{min,j})$$ Now, let $Y=\max_{j=2\ldots n} X_{min,j}$ and $W=X_{1,1}$.

A reminder: if $X_1,\ldots X_n$ are iid with pdf (cdf) $h$ ($H$), then $X_{\min}$ has pdf $h_{\min}=nh(1-H)^{n-1}$ and $X_{\max}$ has pdf $h_{max}=nhH^{n-1}$.
Using this, we get the pdf of $Y$ is $$g(y)=(n-1)mf(1-F)^{m-1}[\int_0^y mf(z)(1-F(z))^{m-1} dz]^{n-2},n\geq 2$$

Note that $Y$ is a statistics that is independent of group 1 so its joint density with any variable in the group 1 is the product of densities.
Now the above probability becomes
$$nm\int_0^x f(w)[\int_0^w \int_y^w f(x_{2,1})dx_{2,1}\ldots\int_y^w f(x_{m,1})dx_{m,1}g(y)dy]dw$$ $$=nm\int_0^x f(w)[\int_0^w (F(w)-F(y))^{m-1}g(y)dy]dw$$ By taking derivative of this integral wrt $x$ and using binomial formula we obtain the pdf of $X_{final}$.

Example: $X$ is uniform, $n=4$, $m=3$. Then
$$g(y)=9(1-y)^2(3y+y^3-3y^2)^2,$$

$$P(X_{final}<x)=(1/55)x^{12}-(12/55)x^{11}$$ $$+ (6/5)x^{10}-(27/7)x^9+(54/7)x^8-(324/35)x^7+(27/5)x^6. $$

Mean of $X_{final}$ is $374/455=0.822$ and its s.d. is $0.145$ .

Answer 2

Since the draws are from an iid samples, we can just consider the draw selected. Consider $f(x) = \frac{d F(x)}{dx}$. Now we know that $b$ is from $f(x)$ and that $b>a$. So,

$$p(b|a) = \frac{f(b)}{\int_a^1 f(y) dy} \forall b>a, 0 \text{ otherwise}.$$

The minimum $m$ in a draw of two is $$p_2(m) = f(m)\int_m^1f(y) dy.$$

The largest minimum among 4 draws would be

$$p(a) = p_2(a)\left[\int_0^a p_2(z) dz\right]^3 = f(a)\int_a^1f(x) dx \left[\int_0^af(y)\left(\int_y^1f(z)dz\right) dy \right]^3.$$

So finally,

$$p(b) = \int_0^1 \left[u(a) \frac{f(b)}{\int_a^1 f(y)dy} f(a)\int_a^1f(x) dx \left[\int_0^af(y)\left(\int_y^1f(z)dz\right) dy \right]^3 \right] da.$$